inequalities,2 solutions

[triclino]

Suppose we were given the following inequality :[math]\frac{x-2}{x}>x+1[/math] ,and the following two solutions were suggested:

 

Solution No 1:

 

 

for x>0 the inequality becomes (after multiplying across):[math]x-2> x^2 +x[/math] ,or [math] 2+x^2<0[/math],but since [math]2+x^2\geq 0[/math] we have a contradiction ,hence the inequality is satisfied for all x<0

 

 

Solution No 2:[math]\frac{x-2}{x}>x+1\Longleftrightarrow\frac{x-2}{x}-(x+1)>0[/math] .

 

And working on the left hand side of the equation we have : [math]\frac{-2-x^2}{x}>0[/math] ,or [math]\frac{2+x^2}{x}<0[/math],and since [math]2+x^2\geq 0[/math] we conclude that x<0 .

 

Hence the inequality is satisfied for all ,x<0

 

Wich of the above solutions is right and which is wrong.Or are they both right,or are they both wrong??

[DrRocket]

Suppose we were given the following inequality :[math]\frac{x-2}{x}>x+1[/math] ,and the following two solutions were suggested:

 

Solution No 1:

 

 

for x>0 the inequality becomes (after multiplying across):[math]x-2> x^2 +x[/math] ,or [math] 2+x^2<0[/math],but since [math]2+x^2\geq 0[/math] we have a contradiction ,hence the inequality is satisfied for all x<0

 

 

Solution No 2:[math]\frac{x-2}{x}>x+1\Longleftrightarrow\frac{x-2}{x}-(x+1)>0[/math] .

 

And working on the left hand side of the equation we have : [math]\frac{-2-x^2}{x}>0[/math] ,or [math]\frac{2+x^2}{x}<0[/math],and since [math]2+x^2\geq 0[/math] we conclude that x<0 .

 

Hence the inequality is satisfied for all ,x<0

 

Wich of the above solutions is right and which is wrong.Or are they both right,or are they both wrong??

 

Your solution is correct, but your logic is invalid in solution 1.

 

In solution 1 you merely showed that no x>0 can satisfy the inequality. You then need to consider separately the case x<0.

 

In solution 2 both the logic and solution are correct. You reduced the problem to showing that a certain fraction which has a negative numerator must be positive, and hence that the denominator, x, must be negative. Since all the operations performed were invertible, the solution set of your final inequality coincides with the solution set of the initial inequality.

[triclino]

Your solution is correct, but your logic is invalid in solution 1..

 

Which law of logic was violated??

[triclino]

Your solution is correct, but your logic is invalid in solution 1.

 

In solution 1 you merely showed that no x>0 can satisfy the inequality. You then need to consider separately the case x<0.

 

 

 

To clarify the case even further ,perhaps ,i must mention the logics behind solution No 1.

 

in solution No1 it was proved that:

 

if x>0 ,then [math] 2+x^2 <0[/math].

 

Now thru the law of contrapositive we can say: [math]\neg(2+x^2)<0\Longrightarrow\neg (x>0)[/math].........................................1

 

But we know that: [math] 2+x^2\geq 0[/math]..............................................................................................................2

 

Also from the law of trichotomy we have: [math]2+x^2\geq 0 \Longrightarrow\neg(2+x^2<0)[/math].......................................................3

 

 

From (1) and (3) and using hypothetical syllogism we can conclude: [math]2+x^2\geq 0\Longrightarrow\neg (x>0)[/math]........................................................4

 

From (2) and (4) and using M ponens we have: [math]\neg(x>0)[/math]...........................................................................................5

 

And since again by trichotomy law we have: x>0 or x<0 ( x=0 is not allowed by the nature of the inequality) and we have proved [math]\neg(x>0)[/math] we can conclude : x<0.

 

As you may very well see we have no need in examining the case x<0